Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 100;
int n, m, q, sz[N], son[N], dfn[N], top[N], tr[N], fa[N], de[N];
int cnt;
int tot, first[N], nxt[N * 2], point[N * 2];
vector<int> p[N];
struct node {
  int a[11], w;
  inline void push(int x) { a[++w] = x; }
} sh[N * 4];
node operator+(node a, node b) {
  node ans;
  ans.w = 0;
  int i = 1, j = 1;
  while (ans.w < 10 && i <= a.w && j <= b.w)
    ans.push((a.a[i] < b.a[j]) ? a.a[i++] : b.a[j++]);
  while (ans.w < 10 && i <= a.w) ans.push(a.a[i++]);
  while (ans.w < 10 && j <= b.w) ans.push(b.a[j++]);
  return ans;
}
inline int read() {
  int f = 1, x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  return x * f;
}
void write(int x) {
  if (x >= 10) write(x / 10);
  putchar('0' + x % 10);
}
inline void add_edge(int x, int y) {
  nxt[++tot] = first[x];
  first[x] = tot;
  point[tot] = y;
}
void dfs(int x, int father) {
  fa[x] = father;
  sz[x] = 1;
  for (int i = first[x]; i != -1; i = nxt[i]) {
    int u = point[i];
    if (u == fa[x]) continue;
    de[u] = de[x] + 1;
    dfs(u, x);
    sz[x] += sz[u];
    if (sz[u] > sz[son[x]]) son[x] = u;
  }
}
void dfs1(int x, int t) {
  top[x] = t;
  dfn[x] = ++cnt;
  tr[cnt] = x;
  if (son[x]) dfs1(son[x], t);
  for (int i = first[x]; i != -1; i = nxt[i]) {
    int u = point[i];
    if (u == fa[x] || u == son[x]) continue;
    dfs1(u, u);
  }
}
inline void pushup(int x) { sh[x] = sh[x + x] + sh[x + x + 1]; }
void build(int x, int l, int r) {
  if (l == r) {
    for (int i = 0; i < (int)p[l].size() && sh[x].w <= 10; i++)
      sh[x].push(p[l][i]);
    return;
  }
  int mid = (l + r) >> 1;
  build(x + x, l, mid);
  build(x + x + 1, mid + 1, r);
  pushup(x);
}
node query(int x, int l, int r, int ll, int rr) {
  if (ll <= l && rr >= r) return sh[x];
  int mid = (l + r) >> 1;
  node ans;
  ans.w = 0;
  if (ll <= mid) ans = ans + query(x + x, l, mid, ll, rr);
  if (rr > mid) ans = ans + query(x + x + 1, mid + 1, r, ll, rr);
  return ans;
}
node query_path(int a, int b) {
  node ans;
  ans.w = 0;
  while (top[a] != top[b]) {
    if (de[top[a]] > de[top[b]]) swap(a, b);
    ans = ans + query(1, 1, n, dfn[top[b]], dfn[b]);
    b = fa[top[b]];
  }
  if (dfn[a] > dfn[b]) swap(a, b);
  ans = ans + query(1, 1, n, dfn[a], dfn[b]);
  return ans;
}
signed main() {
  tot = -1;
  memset(first, -1, sizeof(first));
  n = read();
  m = read();
  q = read();
  for (int i = 1; i < n; i++) {
    int u = read(), v = read();
    add_edge(u, v);
    add_edge(v, u);
  }
  dfs(1, 0);
  dfs1(1, 1);
  for (int i = 1; i <= m; i++) p[dfn[read()]].push_back(i);
  build(1, 1, n);
  node now;
  while (q--) {
    int u = read(), v = read(), a = read();
    now = query_path(u, v);
    int lim = min(a, now.w);
    write(lim), putchar(' ');
    for (int i = 1; i <= min(a, now.w); i++) write(now.a[i]), putchar(' ');
    putchar('\n');
  }
}